Question

An electron is projected at a speed of $2 \times {10^8}m/s$ in horizontal direction along -ve x-axis and a uniform magnetic field of $2 \times {10^{ - 3}}T$ is in vertically upward direction. Find the force on the electron just after it is projected and find the acceleration.

Answer 1

Hint: When charge is at rest only electric field id produce but when charge is under motion then both electric field and magnetic field is produced. Now if there is an existing magnetic field and if a charge is under motion in that existing magnetic field then the magnetic force will be acting on that charge.

Formula used:
$\eqalign{
  & F = qvB\sin \theta \cr
  & F = q\left( {\mathop v\limits^ \to \times \mathop B\limits^ \to } \right) \cr} $
Where
F is the force experienced by the charge
q is the magnitude of charge
v is the velocity of that charge
B is the magnitude of that magnetic field
$\theta $ is the angle between that velocity vector and magnetic field vector

Complete step by step answer:
Let us assume that a charge q is moving with velocity ‘v’ in a magnetic field ‘B’ and the angle between that velocity vector and magnetic field vector is ϴ. Due to that magnetic field there will be a force acting on that moving charge which is $F = qvB\sin \theta $ where q is the magnitude of charge v is the velocity of that charge and B is the magnitude of that magnetic field. Now one has to remember that this force can change only direction of velocity but not magnitude of velocity.
We have the force expression as
$ \Rightarrow F = q\left( {\mathop v\limits^ \to \times \mathop B\limits^ \to } \right)$
From the above expression we can tell that the force is perpendicular to the plane containing both velocity vector and the magnetic field vector.
Anyhow magnitude is given as $F = qvB\sin \theta $
Electron is moving along the negative x axis and magnetic force is in positive z direction i.e upwards. Both are perpendicular hence sinusoidal function will become 1.
$ \Rightarrow \sin \theta = 1$
$ \Rightarrow F = qvB$
We have
$\eqalign{
  & {\text{charge, }}q = - 1.6 \times {10^{ - 19}} \cr
  & {\text{velocity, }}v = 2 \times {10^8}m/s \cr
  & {\text{magnetic field, }}B = 2 \times {10^{ - 3}}T \cr} $
$ \Rightarrow F = 6.4 \times {10^{ - 14}}N$
Mass of electron is $m = 9.1 \times {10^{ - 31}}kg$
Acceleration will be force upon mass
Hence acceleration is
$F = ma$
$\eqalign{
  & \Rightarrow a = \dfrac{F}{m} \cr
  & \Rightarrow a = \dfrac{{6.4 \times {{10}^{ - 14}}N}}{{9.1 \times {{10}^{ - 31}}kg}} \cr
  & \therefore a = 7.03 \times {10^{16}}m/{s^2} \cr} $
Hence both force and acceleration will be directed in negative y direction.

Note:
In the expression where magnetic force is acting on a charge, that force cannot change the magnitude of velocity because that force will be in perpendicular direction to the velocity which means work done by that force is zero which can’t change the kinetic energy of that charge.